I won't swear it's my absolute favorite, but today I learned of a nice proof due to Clementino and TholenClementino and Tholen who take as their starting point the closed-projection characterization of compactness, viz. that a space $X$ is compact iff for every space $Y$, the projection $\pi: Y \times X \to Y$ is a closed map.
If this is assumed, then Tychonoff can be proved without much pain as follows.
Lemma: Let $(X_i)_{i: I}$ be a family of spaces. Then for a point $x$ and subset $A$ of $\prod_{i: I} X_i$, we have $x \in Cl(A)$ (the closure of $A$) if, for every finite $F \subseteq I$, we have $\pi_F(x) \in Cl(\pi_F(A))$ under the projection operator $\pi_F: \prod_{i: I} X_i \to \prod_{i: F} X_i$.
The proof is entirely routine and may be left to the reader.
Proof of Tychonoff: Let $(X_\alpha)_{\alpha \lt \kappa}$ be a family of compact spaces indexed by an ordinal $\kappa$. It is enough to show that the projection
$$Y \times \prod_{\alpha \lt \kappa} X_\alpha \to Y$$
is a closed map for any space $Y$. We do this by induction on $\kappa$. The case $\kappa = 0$ is trivial.
It will be convenient to introduce some notation. For $\gamma \leq \kappa$, let $X^\gamma$ denote the product $Y \times \prod_{\alpha \lt \gamma} X_\alpha$ (so $X^0 = Y$ in this notation), and for $\beta \leq \gamma$ let $\pi_\beta^\gamma: X^\gamma \to X^\beta$ be the obvious projection map. Let $K \subseteq X^\kappa$ be closed, and put $K_\beta := Cl(\pi_{\beta}^\kappa(K))$. In particular $K_\kappa = K$ since $K$ is closed, and we are done if we show $\pi_0^\kappa(K) = K_0$.
Assume as inductive hypothesis that starting with any $x_0 \in K_0$ there is $x_\beta \in K_\beta$ for each $\beta < \kappa$ such that whenever $\beta \lt \gamma \lt \kappa$, the compatibility condition $\pi_\beta^\gamma(x_\gamma) = x_\beta$ holds. In particular, $\pi_0^\beta(x_\beta) = x_0$ for all $\beta \lt \kappa$, and we are now trying to extend this up to $\kappa$.
If $\kappa = \beta + 1$ is a successor ordinal, then the projection
$$\pi_\beta^\kappa: X^\beta \times X_\beta \to X^\beta$$
is a closed map since $X_\beta$ is compact. Thus $\pi_\beta^\kappa(K) = Cl(\pi_\beta^\kappa(K)) = K_\beta$ since $K$ is closed, so there exists $x_\kappa \in K$ with $\pi_\beta^\kappa(x_\kappa) = x_\beta$, and then
$$\pi_0^\kappa(x_\kappa) = \pi_0^\beta \pi_\beta^\kappa (x_\kappa) = \pi_0^\beta(x_\beta) = x_0$$
as desired.
If $\kappa$ is a limit ordinal, then we may regard $X^\kappa$ as the inverse limit of spaces $(X^\beta)_{\beta \lt \kappa}$ with the obvious transition maps $\pi_\beta^\gamma$ between them. Hence the tuple $(x_\beta)_{\beta \lt \kappa}$ defines an element $x_\kappa$ of $X^\kappa$, and all that remains is to check that $x_\kappa \in K$. But since $K$ is closed, the lemma indicates it is sufficient to check that for every finite set $F$ of ordinals below $\kappa$, that $\pi_F(x_\kappa) \in Cl(\pi_F(K))$ (as a subspace of $\prod_{\alpha \in F} X_\alpha$). But for every such $F$ there is some $\beta \lt \kappa$ that dominates all the elements of $F$. One then checks
$$\pi_F(x_\kappa) = \pi_F^\beta \pi_\beta^\kappa(x_\kappa) = \pi_F^\beta(x_\beta) \in \pi_F^\beta(K_\beta) = \pi_F^\beta(Cl(\pi_\beta^\kappa(K))) \subseteq Cl(\pi_F^\beta \pi_\beta^\kappa(K)) = Cl(\pi_F(K))$$
where the inclusion indicated as $\subseteq$ just results from continuity of $\pi_F^\beta$. This completes the proof. $\Box$
(More details at the nLab.)
